無法使用ksoap2解析複雜對象（實體）

我嘗試使用this tutorial和document，但我無法從響應中獲取實體。
這是我的用戶實體：http://pastebin.com/Exx5Fgt6
請求代碼：無法使用ksoap2解析複雜對象（實體）

SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME); 
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11); 

DoubleMarshal marshal = new DoubleMarshal(); 
marshal.register(envelope); 
PropertyInfo userInfo = new PropertyInfo(); 
userInfo.setName("user"); 
userInfo.setValue(user); 
userInfo.setType(user.getClass()); 
request.addProperty(userInfo); 

envelope.setOutputSoapObject(request); 
HttpTransportSE androidHttpTransport = new HttpTransportSE(URL); 
androidHttpTransport.debug = true; 
envelope.addMapping(NAMESPACE, "user", new User().getClass()); 
       androidHttpTransport.call(SOAP_ACTION, envelope); 

Log.i(TAG, "dump Request: " + androidHttpTransport.requestDump); 
Log.i(TAG, "dump Response: " + androidHttpTransport.responseDump); 

SoapObject response = (SoapObject) envelope.bodyIn; 
Log.d(TAG, "Property 0: " +  response.getProperty(0).toString()); 
Log.d(TAG, "Property 1: " + response.getProperty(1).toString()); 
Log.d(TAG, "Property 2: " + response.getProperty(2).toString()); 

// Try pasrsing entity 
User userReponse = (User) envelope.bodyIn; 
Log.d(TAG, "user name: " + userReponse.username);

響應XML：

<SOAP-ENV:Envelope SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:tns="http://axn12.com"> 
<SOAP-ENV:Body> 
    <ns1:registerUserResponse xmlns:ns1="http://axn12.com"> 
     <code xsi:type="xsd:int">1</code> 
     <desc xsi:type="xsd:string">Success. (Executed time: 0.10681s)</desc> 
     <user xsi:type="tns:user"> 
      <user_id xsi:type="xsd:int">5195</user_id> 
      <username xsi:type="xsd:string">xcvx423424c</username> 
      <email xsi:type="xsd:string">[email protected]</email> 
      <password xsi:type="xsd:string">e258314984050bb53a9309592c6f96ab</password> 
      <salt xsi:type="xsd:string">PXN</salt> 
      <id_card xsi:type="xsd:string">Array</id_card> 
      <fullname xsi:type="xsd:string">o0</fullname> 
      <birthdate xsi:nil="true" xsi:type="xsd:string"/> 
      <gender xsi:nil="true" xsi:type="xsd:string"/> 
      <address xsi:nil="true" xsi:type="xsd:string"/> 
      <country xsi:type="xsd:string">233</country> 
      <location xsi:type="xsd:string">0</location> 
      <zipcode xsi:nil="true" xsi:type="xsd:string"/> 
      <mobile xsi:nil="true" xsi:type="xsd:string"/> 
      <is_active xsi:type="xsd:int">0</is_active> 
      <is_lock xsi:type="xsd:int">0</is_lock> 
      <active_token xsi:type="xsd:string">elPp</active_token> 
      <created_date xsi:type="xsd:string">2012-10-11 00:28:36</created_date> 
      <point xsi:type="xsd:double">300000</point> 
      <gold xsi:type="xsd:double">0</gold> 
      <level xsi:type="xsd:int">0</level> 
      <level_point xsi:type="xsd:int">0</level_point> 
     </user> 
    </ns1:registerUserResponse> 
</SOAP-ENV:Body>

雙重元帥：

public class DoubleMarshal implements Marshal { 

public Object readInstance(XmlPullParser parser, String namespace, String name, 
     PropertyInfo expected) throws IOException, XmlPullParserException { 

    return Double.parseDouble(parser.nextText()); 
} 

public void register(SoapSerializationEnvelope cm) { 
    cm.addMapping(cm.xsd, "double", Double.class, this); 

} 

public void writeInstance(XmlSerializer writer, Object obj) throws IOException { 
    writer.text(obj.toString()); 
} 

}

問題是鑄造實體用戶時，我得到錯誤：

10-11 00:08:23.952: W/System.err(1420): java.lang.ClassCastException: org.ksoap2.serialization.SoapObject cannot be cast to com.org.domains.User 
10-11 00:08:23.963: W/System.err(1420): at com.org.ducminh.WebServiceActivty$1.run(WebServiceActivty.java:125) 
10-11 00:08:23.963: W/System.err(1420): at java.lang.Thread.run(Thread.java:856)

哪裏是我的問題嗎？

來源

2012-10-10 R4j

我猜bodiIn包含整個響應。你有沒有嘗試獲得bodyIn作爲SoapObject，然後獲得屬性「user」作爲類User？ – mihail

像魅力一樣工作。非常感謝你：）。請回答我的問題，然後我可以接受它。 – R4j

請回答我的問題，然後我可以接受它。 - R4J

哦，好的:)

的bodyIn包含整個響應。你必須得到它作爲SoapObject然後得到屬性User作爲類User

來源

2012-10-15 08:00:47 mihail

無法使用kso​​ap2解析複雜對象（實體）

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無法使用ksoap2解析複雜對象（實體）